Week 38 (May 12 - 16)
Reading: Atomic physics (Chap. 42)
Key Topics: the Balmer series and atomic spectra, the photoelectric effect, early models of the atom and atomic transitions,
Key Topics: the Balmer series and atomic spectra, the photoelectric effect, early models of the atom and atomic transitions,
Homework Problems:
- Radon gas problem: Radon-222 is a radioactive as produced when radium-226 undergoes alpha decay. If it enters your house through cracks in walls and floors, it may get into the drinking water. Suppose 30 million radon atoms are trapped in the basement. The basement is sealed against further entry of the gas. The half-life of radon is 3.83 days. (a) How many atoms remain after 31 days? Find the activity, dN/dt, (b) just after the basement is sealed and (c) 31 days later. By the way, activity is measured in Becquerels (Bq).
- Skip this problem: Dead sea scrolls problem: The dead sea scrolls, discovered in 1947, were dated by applying radiocarbon dating methods to the linen in which they were wrapped. Linen is made from a flax plant. A Carbon-14 activity of 0.18 Bq per gram of carbon was measured. What was the radiocarbon date of the scrolls?
- Balmer series problem: Find the longest and the shortest wavelength in the Balmer series. You will need to use the Balmer formula, which allows one to find the wavelengths of light emitted by a hydrogen gas. Answer: the longest wavelength (lowest energy photon) corresponds to the transition from n=3 to n=2. Plugging these values into the Balmer formula gives lambda = 656 nm. The shortest wavelength (highest energy photon) corresponds to the transition from n=infinity to n=2. This gives lambda = 364 nm.
- Photon problem: What is the energy of a single photon of red light (wavelength = 632 nm)? Answer: since E = Planck's constant * frequency, we get E = hc/lambda = 3.14 e -19 joules or 1.962 electron-volts.
- Photoelectric effect problem: A 10 milliWatt blue laser (wavelength 400 nm—note that I changed this!) is shone on a gold surface. (i) what is the energy of each blue photon? (ii) how many photons per second strike the surface? (iii) what is the work function of gold (the minimum amount of energy required to knock an electron from the surface) (iii) how much kinetic energy does each emitted electron have? (Let's assume that the photon energy all goes into knocking off electrons, and not heating the gold) (iv) what is the speed (in meters per second) of each emitted electron? Answer: (i) energy = 4.97 e -19 joules or 3.1 eV. (ii) number per second = energy per second / energy per photon = 2 e 16 photons per second. (iii) the work function of gold is 5.31 eV. (iv). So the electrons cannot even be ejected from gold because the photons don't have enough energy to liberate them. If we used potassium, with a work